In FP2 you will learn the method of differences, also known as telescoping sums, for evaluating series.
You will express a sum as another pair of sums by using partial fractions, which will then make it much easier to evaluate the sum.
To illustrate the method of differences, check out the following sums
$$ \begin{align} \sum_{r=1}^{n} \frac{1}{r} - \sum_{r=1}^{n} \frac{1}{r+2} = & \left[ \frac{1}{1}+\frac{1}{2} +\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\ldots+\frac{1}{n} \right] \\ - & \left[ \frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\ldots+\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2} \right] \end{align} $$
What do you notice? There are a whole load of terms in the middle that cancel out. Everything from $\frac{1}{3}$ up to and including $\frac{1}{n}$ is cancelled out. After you're just left with
$$ \begin{align} \sum_{r=1}^{n} \frac{1}{r} - \sum_{r=1}^{n} \frac{1}{r+2} &= 1+\frac{1}{2}+\frac{1}{n+1}+\frac{1}{n+2} \\ &= \frac{3}{2}+\frac{1}{n+1}+\frac{1}{n+2} \end{align} $$
We've just used the method of differences.
In the exam you will be usually given a sum with a summand in the form of a complicated fraction. It'll be in a form like this
$$ \frac{a}{\left(bx+c\right)\left(dx+e\right)} $$
Where $a,b,c,d,e$ are constants.
You will need to split it up into partial fractions so you can use the method of differences. In this section I'll show you some ways you can do this.
This is the method your teacher will probably tell you because it always works.
You set up your answer by writing
$$ \frac{a}{\left(bx+c\right)\left(dx+e\right)} = \frac{\alpha}{\left(bx+c\right)} + \frac{\beta}{\left(dx+e\right)} $$
Then fixing a common denominator
$$ \frac{a}{\left(bx+c\right)\left(dx+e\right)} = \frac{\alpha\left(dx+e\right)+\beta\left(bx+c\right)}{\left(bx+c\right)\left(dx+e\right)} $$
And comparing coefficients of $x$ and the constants in the numerators on either side of the equation
$$ \begin{align} \alpha a + \beta b &= 0 \\ \alpha e + \beta c &= a \end{align} $$
You then solve these simultaneous equations for $\alpha$ and $\beta$.
Q) Express $\frac{2}{\left(x-3\right)\left(x-1\right)}$ in partial fractions.
A) Let
$$ \frac{2}{\left(x-3\right)\left(x-1\right)} = \frac{a}{x-3} + \frac{b}{x-1} $$
Then
$$ \frac{2}{\left(x-3\right)\left(x-1\right)} = \frac{a\left(x-1\right) + b\left(x-3\right)}{\left(x-3\right)\left(x-1\right)} $$
Comparing the numerators
$$ 2 = a\left(x-1\right) + b \left(x-3\right) $$
Comparing coefficients of $x$ and constants we get the simultaneous equations
$$ \begin{align} a + b &= 0 \\ -a -3b &= 2 \end{align} $$
Adding them together, eliminating $a$, yields $b=-1$, therefore $a=1$. Therefore
$$ \frac{2}{\left(x-3\right)\left(x-1\right)} = \frac{1}{x-3} - \frac{1}{x-1} $$
Your teacher will probably balk at this slightly unorthodox method but it's crazy fast and always works for fractions with linear factors. Do not use it for fractions with non-linear factors in the denominator. It does work but it gets complicated which defeats the purpose of using the method.
Once again set up your answer by writing
$$ \frac{a}{\left(bx+c\right)\left(dx+e\right)} = \frac{\alpha}{\left(bx+c\right)} + \frac{\beta}{\left(dx+e\right)} $$
Then look at the denominator of the original fraction on the left. Find the $x$ values that would 'zero' the whole denominator. In this case they are $x=-\frac{c}{b}$ for the first factor and $x=-\frac{e}{d}$ for the second factor.
Then sub in the first value into this fraction. See that we've taken out the corresponding factor $\left(bx+c\right)$
$$ \frac{a}{\left(dx+e\right)} \rightarrow \frac{a}{\left(d\left(-\frac{c}{b}\right)+e\right)} $$
The result of this is $\alpha$.
Once again sub in the second value into this fraction. See that we've taken out the corresponding factor $\left(dx+e\right)$
$$ \frac{a}{\left(bx+c\right)} \rightarrow \frac{a}{\left(b\left(-\frac{e}{d}\right)+c\right)} $$
The result of this is $\beta$.
Q) Express $\frac{2}{\left(x-3\right)\left(x-1\right)}$ in partial fractions.
A) Let
$$ \frac{2}{\left(x-3\right)\left(x-1\right)} = \frac{a}{x-3} + \frac{b}{x-1} $$
Sub $x=3$ into the fraction with the corresponding factor $\left(x-3\right)$ removed to find $a$
$$ a = \frac{2}{\left(3-1\right)} = 1$$
Sub $x=1$ into the fraction with the corresponding factor $\left(x-1\right)$ removed to find $b$
$$ b = \frac{2}{\left(1-3\right)} = -1$$
The Heaviside cover-up method just did the same question in a fraction (ha ha) of the time taken by the standard method. If your teacher doesn't like it tell them it's a legit method.
Now we'll look at some full method of difference example questions.
Q) Evaluate $\sum_{r=1}^n \frac{5}{r\left(r+5\right)}$. Find the limit of the sum as $n \to \infty$.
A) First we'll split up the fraction using partial fractions. Let
$$ \frac{5}{r\left(r+5\right)} = \frac{a}{r} + \frac{b}{r+5} $$
Then
$$ a = \frac{5}{0+5} = 1 $$
And
$$ b = \frac{5}{-5} = -1 $$
Therefore
$$ \frac{5}{r\left(r+5\right)} = \frac{1}{r} - \frac{1}{r+5} $$
So our sum is now $\sum_{r=1}^n \left(\frac{1}{r} - \frac{1}{r+5} \right)$. Evaluating
$$ \begin{align} \sum_{r=1}^n \left(\frac{1}{r} - \frac{1}{r+5} \right) &= \sum_{r=1}^n \frac{1}{r} - \sum_{r=1}^n\frac{1}{r+5} \\ &= \left[ \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\ldots+\frac{1}{n} \right] \\ &- \left[ \frac{1}{5}+\ldots+\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\frac{1}{n+4}+\frac{1}{n+5} \right] \\ &= 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{n+1}-\frac{1}{n+2}-\frac{1}{n+3}-\frac{1}{n+4}-\frac{1}{n+5} \\ &= \frac{25}{12}-\frac{1}{n+1}-\frac{1}{n+2}-\frac{1}{n+3}-\frac{1}{n+4}-\frac{1}{n+5} \end{align} $$
Taking $n \to \infty$, each of the latter fractions with $n$ in the denominator tend to 0, so
$$ \lim_{n \to \infty} \sum_{r=1}^n \frac{5}{r\left(r+5\right)} = \frac{25}{12} $$